Question

Let $c, k \in R$. If $f(x)=(c+1) x^{2}+\left(1-c^{2}\right) x+2 k$ and $f(x+y)=f(x)+f(y)-x y$, for all $x, y \in R$, then the value of $|2(f(1)+f(2)+f(3)+\ldots \ldots+f(20))|$ is equal to _______

Answer 1

$ \begin{array}{l} f(x)=(c+1) x^{2}+\left(1-c^{2}\right) x+2 k \\ \& f(x+y)=f(x)+f(y)-x y \forall x y \in R \\ \lim _{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}=\lim _{y \rightarrow 0} \frac{f(y)-x y}{y} \\ f(x)=-\frac{1}{2} x^{2}+f^{\prime}(0) \cdot x+\lambda \text { but } f(0)=0 \\ \therefore(x)=-\frac{1}{2} x^{2}+\left(1-c^{2}\right) \cdot x \\ \therefore \text { as } f^{\prime}(0)=1-c^{2} \end{array} $
Comparing equation (1) and (2)
We obtain, $c=-3 / 2$
$ \therefore f ( x )=-\frac{1}{2} x ^{2}-\frac{5}{4} x $
Now $\left|2 \sum_{x=1}^{20} f(x)\right|=\sum_{x=1}^{20} x^{2}+\frac{5}{2} \cdot \sum_{x=1}^{20} x$
$ =2870+525 $
$ =3395 $